Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solution() {
  int n, m;
  cin >> n >> m;
  int g[n + 10][m + 10];
  vector<pair<int, int> > row;
  vector<pair<int, int> > col;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      cin >> g[i][j];
    }
  }
  if (n <= m) {
    for (int i = 1; i <= n; ++i) {
      int mn = 1000;
      for (int j = 1; j <= m; ++j) {
        mn = min(mn, g[i][j]);
      }
      if (mn != 0) {
        row.push_back(make_pair(i, mn));
      }
      for (int j = 1; j <= m; ++j) {
        g[i][j] -= mn;
      }
    }
    for (int i = 1; i <= m; ++i) {
      int mn = 1000;
      for (int j = 1; j <= n; ++j) {
        mn = min(mn, g[j][i]);
      }
      if (mn != 0) {
        col.push_back(make_pair(i, mn));
      }
      for (int j = 1; j <= n; ++j) {
        g[j][i] -= mn;
      }
    }
  } else {
    for (int i = 1; i <= m; ++i) {
      int mn = 1000;
      for (int j = 1; j <= n; ++j) {
        mn = min(mn, g[j][i]);
      }
      if (mn != 0) {
        col.push_back(make_pair(i, mn));
      }
      for (int j = 1; j <= n; ++j) {
        g[j][i] -= mn;
      }
    }
    for (int i = 1; i <= n; ++i) {
      int mn = 1000;
      for (int j = 1; j <= m; ++j) {
        mn = min(mn, g[i][j]);
      }
      if (mn != 0) {
        row.push_back(make_pair(i, mn));
      }
      for (int j = 1; j <= m; ++j) {
        g[i][j] -= mn;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (g[i][j] != 0) {
        cout << -1 << endl;
        return;
      }
    }
  }
  int sum = 0;
  for (int i = 0; i < row.size(); ++i) {
    sum += row[i].second;
  }
  for (int i = 0; i < col.size(); ++i) {
    sum += col[i].second;
  }
  cout << sum << endl;
  for (int i = 0; i < row.size(); ++i) {
    int pos = row[i].first;
    int k = row[i].second;
    while (k != 0) {
      cout << "row " << pos << endl;
      k--;
    }
  }
  for (int i = 0; i < col.size(); ++i) {
    int pos = col[i].first;
    int k = col[i].second;
    while (k != 0) {
      cout << "col " << pos << endl;
      k--;
    }
  }
}
int main() {
  int test = 1;
  while (test--) {
    solution();
    cout << endl;
  }
}